# Question related to partial differentiablity and directional derivative

$\mathbf {Question:}$

Define a function $f:\Bbb R^2 \to \Bbb R$ by

$f(x,y)=$ $(x/|y|)\sqrt {x^2+y^2}$ if $y\not = 0$

$f(x,y)=0$ if $y=0$

$\mathbf{a)}$ prove that the function $f$ is not continuous at the point $(0,0)$

$\mathbf{answer-a:}$

I need to find a sequence $\{u_k\}_{k\in \Bbb N}$ converges to $(0,0)$ such that $\{f(u_k)\}$ does not converge to $f(u)$

But I could not find such a sequence. Please can someone give me a hint about the sequence?

$\mathbf{b)}$ The function f has directional derivatives in all directions at the point $(0,0)$

$\mathbf{answer-b:}$ I need to prove that $\frac{\partial f}{\partial p} (0,0)=\,lim_{t\to 0}\frac{f((0,0)+tp)-f(0,0)}{t}$ exists.

Is this right? If this is true, please show me how to prove its existence?

$\mathbf{c)}$ prove that if $c$ is any number, then there is a vector $p$ of norm $1$ such that $\frac{\partial f}{\partial p} (0,0)=c$

$\mathbf{answer-c:}$ I could not produce any idea to solve the part.

Please can someone show and explain me my questions step by step? I am just starting real analysis and on my own. So I am confused so much :( thank you for helping :)

Hint

a) Take $(u_k)=(\frac{1}{k},\frac{1}{k^3})$.

b) If $p=(p_1,p_2)$ such that $p_2\neq 0$ $$\frac{\partial f}{\partial p} (0,0)=\lim_{t\to 0}\frac{f((0,0)+tp)-f(0,0)}{t}=\lim_{t\to 0}\frac{f((tp_1,tp_2))}{t}=\lim_{t\to 0}\frac{tp_1\sqrt{(tp_1)^2+(tp_2)^2}}{t|tp_2|}=\frac{p_1\sqrt{(p_1)^2+(p_2)^2}}{|p_2|}$$

and if $p=(p_1,0)$ then it's clear that $\frac{\partial f}{\partial p} (0,0)=0$

c) If $c=0$ take $p=(1,0)$ and if $c\neq 0$ then $$\frac{\partial f}{\partial p} (0,0)=\frac{p_1}{|p_2|}=c$$ and then $p_1=c|p_2|$ and since $p_1^2+p_2^2=c^2p_2^2+p_2^2=1$ then $$p=\left(\frac{c}{\sqrt{1+c^2}},\frac{1}{\sqrt{1+c^2}}\right)$$

• $(u_k)\to 0$ as $k\to \infty$ okay! But $\{f(u_k)\} \to 0$ so f is continuous at (0,0) but I dont want f to be continuous here – B11b Jul 7 '13 at 1:08
• $f(1/k,1/k)=(1/k)/(1/k\sqrt{2/k^2})=k^2/k\sqrt{2}\to\infty$ – user63181 Jul 7 '13 at 1:13
• Aaow! I took the limit false:( sorry:( by the way, can you help me for part b and c? Please.. – B11b Jul 7 '13 at 1:15
• Ok so I'll edit my answer. – user63181 Jul 7 '13 at 1:29
• Enjoy with the answer and good luck. – user63181 Jul 7 '13 at 1:46